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0580/43 May/Jun 2023, Question 6: Estimating the Mean for Grouped Data

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 15 marks

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The question, in short

From a cumulative frequency table of 200 car speeds, draw the curve and estimate median, interquartile range and count above 110 km/h; then estimate mean truck mass and histogram block heights from a frequency table.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (ai) · 3 marks

Plot cumulative frequency against speed on the grid and join points to draw the cumulative frequency diagram.

  1. 1Plot each cumulative frequency at the upper class boundary
  2. 2Points (70, 12), (80, 46), (90, 115), (95, 155), (100, 177), (120, 200)
  3. 3Join them with a smooth increasing curve

AnswerCorrect increasing curve through (70, 12), (80, 46), (90, 115), (95, 155), (100, 177), (120, 200)

Part (aiia) · 1 mark

Read the median speed from your cumulative frequency diagram.

  1. 1Median at cumulative frequency 200 ÷ 2 = 100
  2. 2Read across from 100 to the curve and down: about 88 km/h

Answer87 to 89.5

Part (aiib) · 2 marks

Find the interquartile range of speeds using your cumulative frequency diagram.

  1. 1Lower quartile at 50: about 81 km/h
  2. 2Upper quartile at 150: about 94.2 km/h
  3. 3IQR = 94.2 − 81 = 13.2 km/h

Answer12.5 to 14

Part (aiic) · 2 marks

Estimate how many cars have a speed greater than 110 km/h from your diagram.

  1. 1Read the curve at v = 110: cumulative frequency about 189
  2. 2Cars faster than 110 km/h = 200 − 189 = 11

Answer10 (accept 6 to 13)

Part (bi) · 4 marks

Calculate an estimate of the mean mass of the 50 trucks using midpoints of the mass classes.

  1. 1Midpoints: 2300, 3050, 4250, 5350
  2. 2Σfx = 12×2300 + 15×3050 + 16×4250 + 7×5350
  3. 3= 27 600 + 45 750 + 68 000 + 37 450 = 178 800
  4. 4Mean = 178 800 ÷ 50 = 3576 kg

Answer3576

Part (bii) · 3 marks

Given the first histogram block is 6 cm high, calculate the heights of the other three blocks.

  1. 1First block: 12 ÷ 600 = 0.02, drawn 6 cm high, so height = frequency density × 300
  2. 215 ÷ 900 × 300 = 5 cm
  3. 316 ÷ 1500 × 300 = 3.2 cm
  4. 47 ÷ 700 × 300 = 3 cm

Answer5, 3.2, 3

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Also searched as: 0580/43/M/J/23 · 0580_s23_qp_43 · 0580 s23 43 · question 6.
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